Question

A uniform cube of mass M and side a is placed on a frictionless horizontal surface. A vertical force F is applied to edge as shown in figure.

Match Column $I$ with Column $II$.

Column I		Column I
A	$\frac{Mg}{4}<F<\frac{Mg}{2}$	p	Cube will move up
B	$F>\frac{Mg}{2}$	q	Cube will not exhibit motion
C	$F>Mg$	r	Cube will begin to rotate and slip at A
D	$F=\frac{Mg}{4}$	s	Normal reaction effectively at $a/3$ from $A$ no motion

• A. $A-p,B-q,C-s,D-r$
• B. $A-r,B-s,C-q,D-p$
• C. $A-q,B-r,C-p,D-s$
• D. $A-s,B-p,C-r,D-q$

Answer 1

Answer: (C)

From figure, Moment of force $F$ about $A$,
${\tau }_1=F\times a$, anticlockwise.
Moment of weight $Mg$ of cube about $A$,
${\tau }_2=Mg\times \frac{a}{2}$, clockwise.
The cube will not exhibit any motion,
if ${\tau }_1={\tau }_2$
or $F\times a=Mg\times \frac{a}{2}$ or $F=\frac{Mg}{2}$
The cube will rotate only, when ${\tau }_1>{\tau }_2$
$F\times a>Mg\frac{a}{2}$ or $F>\frac{Mg}{2}$
If we assume that normal reaction is effective at $a/3$ from $A$, then block would turn if
$Mg\times \frac{a}{3}=F\times a$ or $F=\frac{Mg}{3}$.
When $F=\frac{Mg}{4}<\frac{Mg}{3}$, there will be no motion.
Hence, we conclude $A-q;B-r;C-p;D-s$.